Prove that if $a,b,c\in \mathbb{R}$ are all distinct, then $a+b+c=0$ if and only if $(a,{a}^{3}),(b,{b}^{3}),(c,{c}^{3})$ are collinear.

Kaiden Wilkins
2022-05-07
Answered

Prove that if $a,b,c\in \mathbb{R}$ are all distinct, then $a+b+c=0$ if and only if $(a,{a}^{3}),(b,{b}^{3}),(c,{c}^{3})$ are collinear.

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Ariella Bruce

Answered 2022-05-08
Author has **12** answers

Step 1

The first part of your proof is enough because you can proceed by equivalence

$\text{aligned points}\text{}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\text{}\text{equal slopes}\text{}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\text{}{c}^{2}+bc+{b}^{2}={b}^{2}+ab+{a}^{2}\text{}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\text{}(a-c)(a+b+c)=0\text{}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\text{}(a+b+c)=0$

Here is an alternate proof:

Using the classical alignment criteria for 3 points $({x}_{k},{y}_{k})$ for k=1,2,3 which is

$\left|\begin{array}{ccc}{x}_{1}& {x}_{2}& {x}_{3}\\ {y}_{1}& {y}_{2}& {y}_{3}\\ 1& 1& 1\end{array}\right|=0$

This means that we have to show that

$\left|\begin{array}{ccc}a& b& c\\ {a}^{3}& {b}^{3}& {c}^{3}\\ 1& 1& 1\end{array}\right|=0\text{}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\text{}a+b+c=0$

But this is very easy because the determinant can be factorized in the following way:

$(a-c)(b-a)(b-c)(a+b+c),$

knowing that a,b,c are all different.

In fact, I just realized that I had already answered a similar question here... with two proofs, this one and another one based on a third degree equation with no term in ${x}^{2}$

The first part of your proof is enough because you can proceed by equivalence

$\text{aligned points}\text{}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\text{}\text{equal slopes}\text{}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\text{}{c}^{2}+bc+{b}^{2}={b}^{2}+ab+{a}^{2}\text{}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\text{}(a-c)(a+b+c)=0\text{}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\text{}(a+b+c)=0$

Here is an alternate proof:

Using the classical alignment criteria for 3 points $({x}_{k},{y}_{k})$ for k=1,2,3 which is

$\left|\begin{array}{ccc}{x}_{1}& {x}_{2}& {x}_{3}\\ {y}_{1}& {y}_{2}& {y}_{3}\\ 1& 1& 1\end{array}\right|=0$

This means that we have to show that

$\left|\begin{array}{ccc}a& b& c\\ {a}^{3}& {b}^{3}& {c}^{3}\\ 1& 1& 1\end{array}\right|=0\text{}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\text{}a+b+c=0$

But this is very easy because the determinant can be factorized in the following way:

$(a-c)(b-a)(b-c)(a+b+c),$

knowing that a,b,c are all different.

In fact, I just realized that I had already answered a similar question here... with two proofs, this one and another one based on a third degree equation with no term in ${x}^{2}$

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